# 11b. Scalar and vector products

The following questions illustrate the key concepts needed in this topic. See if you can answer them.
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It is given that vectors $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \mathbf{b} = \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix}, \mathbf{c} = \begin{pmatrix} -1 \\ 1 \\ z \end{pmatrix}$ and $\mathbf{d} = \begin{pmatrix} k+2 \\ k-1 \\ k \end{pmatrix}$ where $z$ and $k$ are constants.

Points $A$ and $B$ have coordinates $A(1,2,-3)$ and $B(5,0,1)$.

Points $A$ and $B$ have coordinates $A(-1,2,3)$ and $B(5,0,1)$.
Remark: $A$ has changed compared to the earlier questions.

It is given that $\mathbf{a} = -\mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}$.

It is given that the direction cosines of a vector $\mathbf{b}$ with respect to the $x$ -and $y$- axes are $\frac{2}{\sqrt{6}}$ and $\frac{-1}{\sqrt{6}}$ respectively.

Remark: For the following questions, do not use $\mathbf{a}$ and $\mathbf{b}$ from the previous questions. Vectors $\mathbf{a}$ and $\mathbf{b}$ are now unknown.

With respect to the origin $O$, points $A,B$ and $C$ have position vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}=3\mathbf{a}+\mathbf{b}$ respectively.

It is further given that $|\mathbf{a}| = 2, \angle AOB = 60^\circ$ and $\mathbf{b}$ is a unit vector.

Points $A,B$ and $C$ have position vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ respectively.